91Ó°ÊÓ

This state is when the solution has reached a point in which no more solute can be added. The addition of solute after this point would result in a solid precipitate or gas being released. Such a mixture is called a saturated solution

Reaction of

Ca(OH)₂(s) ⇌ Ca²⁺ (aq) +2OH^- (aq)

Reaction of

K_spof Ca(OH)₂is K_sp = [Ca²⁺] . [OH^-] = 1.3.10^-6

The molar solubility of Ca²⁺ is ‘s’

The molar solubility of OH^-is ‘2s’

Let us find the value of

\begin{aligned}{{K_{sp}}=\left[{C{a^{2+}}}\right]\cdot{{\left[{O{H^-}}\right]}^2}}\\{1.3\cdot{{10}^{-6}}=s\cdot{{(2s)}^2}}\\{4{s^3}=1.3\cdot{{10}^{-6}}}\\{s=6.875\cdot{{10}^{-3}}}\\{\left[{O{H^-}}\right]=2s=1.375\cdot{{10}^{-2}}{\rm{M}}}\end{aligned}

Let us calculate

\begin{aligned}{pOH=-\log\left[{O{H^-}}\right]}\\{=-\log\left[{1.375\cdot{{10}^{-2}}}\right]}\\{=1.86}\end{aligned}

pH Of the saturated solution

\begin{aligned}{pH=14-pOH}\\{=14-1.86}\\{=12.14}\end{aligned}